Re: evenly distributed subset of points on a sphere
- From: "Abel Brown" <brown.2179@xxxxxxx>
- Date: Fri, 30 Oct 2009 17:47:03 +0000 (UTC)
ImageAnalyst <imageanalyst@xxxxxxxxxxxxxx> wrote in message <067a12b6-b55b-40b8-8c9c-474d70729e13@xxxxxxxxxxxxxxxxxxxxxxxxxxx>...
On Oct 30, 12:09?pm, "Abel Brown" <brown.2...@xxxxxxx> wrote:
LOL, I did not generate this set of 250 points. ?
This is a little strange. ?You seem to be avoiding the question. I need a practical solution not a mathematical proof. ?
250 evenly distributed around the earth. ?Need to divide these 250 points into 6 subsets or subnetworks of ~ 250/6 = 40 points each which are themselves evenly distributed around the earth.
Indeed, generate 40 evenly distributed points on a sphere and choose closes points in super set. ?Easy enough ... but how to generate 40 evenly distributed points around the earth?
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Abel:
I would imagine that you'd just take every 6th point in the Z
direction. Visualize this: you have a bunch (say 250) of points more
or less randomly distributed around the surface of the globe. Now
imagine that you take a string and fasten it to the point closest to
the North Pole. Now spiral that string around the globe such that it
traverses every point, say as the string is going from east to west as
you wind the string around the globe clockwise (looking down from the
top). So now every point on the globe is ordered as to its location
on the string. Now assuming your initial set was fairly randomly and
uniformly distributed, I think you could get one set which consists of
points
1, 7, 13, 19, 25, etc.
The second set would be one over from that:
2, 8, 14, 20, 26, etc.
and the third set would consist of points
3, 9, 15, 21, 27, etc.
and so on. Imagine that each set was identified by different colored
beads on the surface of the globe. It seems like the set of (say) red
beads (taken starting with location 1) would be approximately just as
uniformly scattered as the (say) green beads (take starting with
location 2), and so on. Every set of different colored beads would be
about as randomly and uniformly positioned as any other set.
Because we picked up the points by spiraling down, subsequent points
have decreasing Z values (latitude). So all you need to do is to sort
your array of x,y,z coordinates by the z value and then take every nth
point. Other sets just start at a different point before taking every
nth point, like my example above. Does that make sense - do you
follow me? Does that seem like it will work well enough for you?
Might be worth a try and see how it goes.
Regards,
ImageAnalyst
Brilliant! I really like this idea! Nice and simple. I'll give it a try and let you know how it works out.
Fetterman: Thanks for your reply! Indeed, I did some reading/googling before posting and pretty much all of the algorithms I found were overkill. Many folks on the forum have tremendous practical experience. It's always interesting to see what solutions come up!
Thanks again!
.
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- From: Abel Brown
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